maximize: $xy-x$ and s.t. $x^2+y^2 \leq 9$ and $x \ge 0$

Then, I tried to find points using lagrange method. I used 4 cases and they are $(1)$ $\lambda_1 =0$ and $\lambda_2 = 0$



Maximize $xy - x$, given constraints: $x^2 + y^2 \leq 9, x \ge 0$

Taking $xy - x = \lambda (x^2 + y^2 -9)$

Taking derivative wrt $x$ and $y$, we get -

$y-1 = 2 \lambda x$ ...(i)

$x = 2 \lambda y$ ...(ii)

From (ii), $\lambda = \frac{x}{2y}$ and substituting in (i)

we get $y(y-1) = x^2 \implies y(y-1) + y^2 \leq 9$

That gives, $y^2 - \frac{y}{2} \leq \frac{9}{2} \implies (y - \frac{1}{4})^2 \leq \frac{73}{16}$

So $y \leq \frac{\sqrt 73 + 1}{4} \approx 2.386001\, $ (only taking positive values as $x \ge 0$ and we need $xy$ to be positive for max value)

taking max value of $y$, gives $x \approx 1.818516$

And max value of $xy-x \approx 2.520465$

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    $\begingroup$ @Jumbo09 I did not use $\lambda_1$ and $\lambda_2$ and both constraints as it would complicate it. $x \geq 0$ is a check I did when deciding which values to consider$. $\endgroup$ – Math Lover Oct 21 '20 at 8:56
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    $\begingroup$ @Jumbo09 In other words if the solution gave us certain values where $x$ was negative, I would ignore those negative one's instead of adding it to the constraint upfront. $\endgroup$ – Math Lover Oct 21 '20 at 9:07
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    $\begingroup$ ,so Which points are failed ? I didn't get it $\endgroup$ – Jumbo09 Oct 21 '20 at 9:19
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    $\begingroup$ I understood this part , but I mean that when I solve this optimization problem, Points which are failed on constraint qualification become the candidate. Therefore, for this fail, what should I write the candidates in order to find the maximum. $\endgroup$ – Jumbo09 Oct 21 '20 at 9:35
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    $\begingroup$ @Jumbo09 ok I understand you now. Your case 2, $\lambda_2 = 0, \lambda_1 \geq 0$ should have given your maximum. Please check where you did a mistake in calculation. That case is very similar to what my solution is. $\endgroup$ – Math Lover Oct 21 '20 at 9:42

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